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But   L = £
.>'      ^

Appendix IL

. ._»_



and the polar distance is fixed. Taking base D M, summate ABC
from pole o, obtaining D E F M, shewing total load on the left of
any vertical. Next make D H = M F, and with H as pole summate
D E F to base D j, obtaining D G. A vertical through G will pass

through the centre-of-gravity of the load. Set off a c^ A c, ag^ A G,
and draws£- K horizontally. Then D L and F K are right and left
reactions respectively. The deduction of area DLKM from
D E F M gives the two triangular areas, whose verticals indicate
shear, or total load on left of section less left reaction.